Glycerine flows steadily through a horizontal tue of length 1.5 m and radius 1.0 cm. if the amount of glycerine collected per second at one end is `4.0xx10^(-3)kgs^(-1)`, what is the pressure difference between the two ends of the tube? (Density of glycerine `=1.3xx10^(3)kgm^(-3)` and viscosity of glycerine `=0.83Pa` s). [you may also like to check if the assumption of laminar flow in the tube is correct].

Options

A. 64 Pa
B. 640 Pa
C. 6400 Pa
D. ` 10 ^5 ` Pa

Answer

B

Explanation

Given ` l = 1.2 m, r = 1 cm = 1 xx 10 ^ (-2) m, eta = 0.8 Nsm ^ ( -2) `
Volume of glycerine flowing per second is `Q = (3.14 xx 10 ^(-3) kg s ^ ( -1))/(1.2 xx 10 ^(3) kg m ^ ( -3)) = (3.14 xx 10 ^(-6))/ (1.2 ) m ^ 3 s ^(-1) `
From Poisseuille’s formula, ` Q = ( pi (P _ 1 - P _ 2 ) r ^4 ) / ( 8 eta l ) ` We have ` Deltap = ( 8 eta l Q ) /( pi r ^4 ) `
Substituting the above values, we get ` Delta p = 6.4 xx 10 ^(2) Pa`.